$A$ parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$. What will be the capacitance if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

$A$ capacitor with plate separation $d$ is charged to $V$ volts. The battery is disconnected and a dielectric slab of thickness $\frac{d}{2}$ and dielectric constant $K=2$ is inserted between the plates. The potential difference across its terminals becomes

$A$ parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The initial level of the liquid is $\frac{d}{3}$. Suppose the liquid level decreases at a constant speed $V$. Find the time constant $\tau$ as a function of time $t$.

The capacitance of an air-filled parallel plate capacitor is $10 \, pF$. The separation between the plates is doubled and the space between the plates is then filled with wax,giving the capacitance a new value of $40 \times 10^{-12} \, F$. The dielectric constant of wax is:

$A$ parallel plate capacitor having capacity $C_0$ is charged to $V_0$. With the battery disconnected,if the separation between the plates is doubled,then the energy stored in it is $E_1$. Instead,if the separation between the plates is doubled with the battery in connection,the energy stored in it is $E_2$. Then the value of $\frac{E_2}{E_1}$ is

$A$ parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$,its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is